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August 5, 2011 / ryanbtaylor

Easy way to work out which machine is on which subnet.

OK so imagine the scene, you’re taking your exam (Microsoft, Cisco etc…) and you get a networking question, something along the lines of the question below:

You work in an international company which is named Wiikigo. Before entering this company, you have two years of experience in the IT field, as well as experience implementing and administering any Windows client operating system in a networked environment. You are professional in installing, upgrading and migrating to Windows 7, deploying Windows 7, and configuring Hardware and Applications and so on. There is a head office and a branch office in your company. The exhibit below shows the configuration of the relevant portion of the network. In the branch office, you deploy a new computer named C01 that runs Windows 7. You need to assign an IP address to C01.

What IP Address should you use?

A. 192.168.2.63

B. 192.168.2.65

C. 192.168.2.30

D. 192.168.2.40

Before we can use a quick calculation to work out the correct IP for the subnet we need to convert the CIDR notation of the subnet to decimal, we know from the above example the the subnet mask is 27 bit so in binary looks it like this >>> 11111111 11111111 11111111 11100000.  Once we have converted the subnet mask to decimal it looks like this >>> 255.255.255.224.  Ok now we’ve converted the subnet mask and we know that we’re using the forth octet to split the network we can now use the trusty AND function on a scientific calcuator (which you can use in your Microsoft or Cisco exams.)

So starting with the routers ip address as a reference (remember we are only internested in the 4th octet):

Router: 62 AND 224 = 32, the router is on a network with a networkID on 32

A: 63 AND 224 = 32, this IP address is on networkID 32 but may not be a valid IP on the network, you’ll see why in a minute.

B. 65 AND 224 = 64, this IP address is on networkID 64, this IP address would not be valid.

C. 30 AND 224 = 0, this IP address is on networkID 0, this IP address would not be valid.

D. 40 AND 224 = 32,  this IP address is on networkID 32 but may not be a valid IP on the network, you’ll see why in a minute.

OK so using the above method we’ve been able to rule out 2 IP addresses as invalid, leaving is with 192.168.2.63 & 192.168.2.40, are both of these IPs valid?  No, lets look why.  Again if we focus on the forth octet of these 2 IP address and covert them both to binary we’ll see why we can’t use one these IPs.

.63 = .11111111  <<< This is the all 1’s address (remember we lose two IPs from the subnet the all 0’s address and the all 1’s address.)

.40 = .00101000 <<< This is the valid address we’re after.

So the answer to the above question is D.

The above example was quite a tricky question which was why I chose it as a example.  This post assumes that you understand subnetting but if you don’t or are a little confused stay tuned for a post on subnetting in more details.  For more information on the AND function please follow the link.  I would like to take this opportunity to thank Lee Holland who very kindly took a massive amount of time out to teach me subnetting many moons ago.

Thank you for reading I hope you find the information useful, feel free to join in using comments section below.

-Ryan Taylor

 

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